Đáp án:

$\begin{array}{l}
b)C = x\left( {x + 1} \right)\left( {x + 5} \right)\left( {x + 4} \right)\\
 = x\left( {x + 5} \right)\left( {x + 1} \right)\left( {x + 4} \right)\\
 = \left( {{x^2} + 5x} \right)\left( {{x^2} + 5x + 4} \right)\\
 = {\left( {{x^2} + 5x} \right)^2} + 4.\left( {{x^2} + 5x} \right)\\
 = {\left( {{x^2} + 5x} \right)^2} + 4.\left( {{x^2} + 5x} \right) + 4 - 4\\
 = {\left( {{x^2} + 5x + 2} \right)^2} - 4 \ge  - 4\\
 \Rightarrow C \ge  - 4\\
 \Rightarrow GTNN:C =  - 4\\
Khi:{x^2} + 5x + 2 = 0\\
 \Rightarrow {x^2} + 2.x.\dfrac{5}{2} + \dfrac{{25}}{4} = \dfrac{{21}}{4}\\
 \Rightarrow {\left( {x + \dfrac{5}{2}} \right)^2} = \dfrac{{21}}{4}\\
 \Rightarrow x = \dfrac{{ - 5 \pm \sqrt {21} }}{2}\\
c)B = 5{x^2} + 5{y^2} + 8xy + 2y - 2x + 2020\\
 = \left( {4{x^2} + 8xy + 4{y^2}} \right) + \left( {{x^2} - 2x + 1} \right)\\
 + \left( {{y^2} + 2y + 1} \right) + 2018\\
 = 4{\left( {x + y} \right)^2} + {\left( {x - 1} \right)^2} + {\left( {y + 1} \right)^2} + 2018\\
 \Rightarrow B \ge 2018\\
 \Rightarrow GTNN:B = 2018\\
Khi:x = 1;y =  - 1\\
d)D = {x^2} - 8y + 5x + {y^2} + 18\\
 = {x^2} + 5x + \dfrac{{25}}{4} + {y^2} - 8y + 16 - \dfrac{{21}}{4}\\
 = {\left( {x + \dfrac{5}{2}} \right)^2} + {\left( {y - 4} \right)^2} - \dfrac{{21}}{4} \ge \dfrac{{ - 21}}{4}\\
 \Rightarrow D \ge  - \dfrac{{21}}{4}\\
 \Rightarrow GTNN:D =  - \dfrac{{21}}{4}\\
Khi:\left\{ \begin{array}{l}
x = \dfrac{{ - 5}}{2}\\
y = 4
\end{array} \right.
\end{array}$